Assume that population mean is to be estimated from the sample described. Use the sample results to approximate the margin of error and $95 \%$ confidence interval. Sample size, $n=100$; sample mean, $\bar{x}=75.0 \mathrm{~cm}$; sample standard deviation, $s=5.0 \mathrm{~cm}$ The margin of error is $\square \mathrm{cm}$. (Round to one decimal place as needed.) Find the $95 \%$ confidence interval. $\square \mathrm{cm}<\mu<\square \mathrm{cm}$ (Round to one decimal place as needed.)

Step 1 ：Given that the sample size, \(n=100\); the sample mean, \(\bar{x}=75.0 \mathrm{~cm}\); and the sample standard deviation, \(s=5.0 \mathrm{~cm}\).

Step 2 ：We are to assume that the population mean is to be estimated from the sample described and use the sample results to approximate the margin of error and the 95% confidence interval.

Step 3 ：The margin of error, \(E\), can be calculated using the formula \(E = Z \times \frac{s}{\sqrt{n}}\), where \(Z\) is the Z-score corresponding to the desired confidence level. For a 95% confidence level, \(Z = 1.96\).

Step 4 ：Substituting the given values into the formula, we get \(E = 1.96 \times \frac{5.0}{\sqrt{100}} = 0.98 \mathrm{~cm}\).

Step 5 ：The 95% confidence interval can be calculated using the formula \(\bar{x} - E < \mu < \bar{x} + E\).

Step 6 ：Substituting the given values into the formula, we get \(75.0 - 0.98 < \mu < 75.0 + 0.98\), which simplifies to \(74.02 \mathrm{~cm} < \mu < 75.98 \mathrm{~cm}\).

Step 7 ：Final Answer: The margin of error is \(\boxed{0.98 \mathrm{~cm}}\). The 95% confidence interval is \(\boxed{74.02 \mathrm{~cm}<\mu<75.98 \mathrm{~cm}}\).